Let $w_1,...,w_r ∈ C^n$ be arbitrary orthogonal vectors. The orthogonal projector onto the complement of the subspace spanned by the $w_i$ is $P = I − \sum_{i=1}^{r} \frac{w_iw_i^*}{w_i^{*}w_i}$. Find an SVD for $P$, that is, unitary matrices $U$, $V$ and a diagonal $\Sigma$ such that $P = U \Sigma V^*$.
My first step was to find the eigenvalues and eigenvectors of $P^*P$. Since $P$ is an orthogonal projector, we have that $P^* = P$ and $P^2 = P$. So finding the eigenvalues and vectors of $P^*P$ is equivalent to finding them for $P$. So if we let the subspace $A = span\{w_1,...w_r\}$, and $B$ be the complement of A, then for every $x \in C^n$, we can express x as $$ x = x_A + x_B $$ with $x_A \in A$ and $x_B \in B$. So we have $$ Px = P(x_A + x_B) $$ After this step I'm a little unsure of what to do next. Can I say the following: $$ P(x_A + x_B) = 0 + x_B $$, which tells me that $Px_B = x_B$ and so the eigenvalues for $P$ are simply 1? Also I'm not sure what to think of $x_B$. Any help would be greatly appreciated.
Let $M$ be a subspace of $\mathbb{C}^{n}$. Then $Px$ is the orthogonal projection of $x$ onto $M$ iff $$ (x-Px)\perp M,\;\;\; Px \in M. $$ And $Qx$ is the orthogonal projection of $x$ onto $M^{\perp}$ iff $$ (x-Qx)\perp M^{\perp},\;\;\; Qx \in M^{\perp}. $$ Notice that $$ (x-(x-Px))\perp M^{\perp},\;\;\; x-Px \in M^{\perp}. $$ Therefore $Qx=(I-P)x$.